Today I encounter sometihing strange, at least to me in elementary linear algebra.

Suppose that there are two real symmetric matrices, and is definite positive. Then there is a square matrix such that where is the transpose of .

At first glance, I think of diagonilization in linear algebra. And so this result strike me: we can always diagonalize two symmetric matrices simultaneuously!

Then searching on the Internet for a while, I found that I was wrong. This is not diagonalization. This is related more to (symmetric) bilinear forms than to linear transformations. This is a very important difference.

Having found the starting point, things become easier to understand. We can give with the standard inner product and define . This is a symmetric bilinear form, and is non-degenerate since is not. So there is an orthogonal basis for , that is to say, under this basis , looks like the standard inner product, . And thus with the matrix , we have that . We can thus rewrite the bilinear form as . So, under this expression, we can define another bilnear form, . So, since is symmeytic, we can find an orthonormal basis, such that under this new basis, we have . We denote the orthogonal matrix as .

All having been said, let’s look at what happens here. in the basis , is the usual inner product. So we have that . Under the basis, , becomes a diagonal matrix , so we have that . Note that we also have . So at last we find a matrix such that and .

One thing to keep in mind if there is something worth to do so: **bilinear form and linear transformation are different**.

Note that the result nevers says that two symmetric matrices can be simultaneuouly diagonalized even if there is at least one of them is definite positive. This will mean that they commute with each other. But it is not the case in general. we can found some very simple counterexamples.

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